JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Spring Pendulum

  • question_answer
    A mass m is suspended separately by two different springs of spring constant K1 and K2 gives the time-period \[{{t}_{1}}\] and \[{{t}_{2}}\]respectively. If same mass m is connected by both springs as shown in figure then time-period t is given by the relation                    [CBSE PMT 2002]

    A)            \[t={{t}_{1}}+{{t}_{2}}\]

    B)            \[t=\frac{{{t}_{1}}.{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\]

    C)            \[{{t}^{2}}={{t}_{1}}^{2}+{{t}_{2}}^{2}\]

    D)            \[{{t}^{-2}}={{t}_{1}}^{-2}+{{t}_{2}}^{-2}\]

    Correct Answer: D

    Solution :

                       \[{{t}_{1}}=2\pi \sqrt{\frac{m}{{{K}_{1}}}}\]and \[{{t}_{2}}=2\pi \sqrt{\frac{m}{{{K}_{2}}}}\] Equivalent spring constant for shown combination is K1 + K2. So time period t is given by \[t=2\pi \sqrt{\frac{m}{{{K}_{1}}+{{K}_{2}}}}\] By solving these equations we get \[{{t}^{-2}}=t_{1}^{-2}+t_{2}^{-2}\]


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